HDU 6140 Hybrid Crystals【题意】【思维】

题目链接：

2017 Multi-University Training Contest - Team 8 - H

Problem Description

Kyber crystals, also called the living crystal or simply the kyber, and known as kaiburr crystals in ancient times, were rare, Force-attuned crystals that grew in nature and were found on scattered planets across the galaxy. They were used by the Jedi and the Sith in the construction of their lightsabers. As part of Jedi training, younglings were sent to the Crystal Caves of the ice planet of Ilum to mine crystals in order to construct their own lightsabers. The crystal’s mix of unique lustre was called “the water of the kyber” by the Jedi. There were also larger, rarer crystals of great power and that, according to legends, were used at the heart of ancient superweapons by the Sith.

— Wookieepedia

Powerful, the Kyber crystals are. Even more powerful, the Kyber crystals get combined together. Powered by the Kyber crystals, the main weapon of the Death Star is, having the firepower of thousands of Star Destroyers.

Combining Kyber crystals is not an easy task. The combination should have a specific level of energy to be stablized. Your task is to develop a Droid program to combine Kyber crystals.

Each crystal has its level of energy (i-th crystal has an energy level of ai). Each crystal is attuned to a particular side of the force, either the Light or the Dark. Light crystals emit positive energies, while dark crystals emit negative energies. In particular,

• For a light-side crystal of energy level ai, it emits +ai units of energy.
• For a dark-side crystal of energy level ai, it emits −ai units of energy.

Surprisingly, there are rare neutral crystals that can be tuned to either dark or light side. Once used, it emits either +ai or −ai units of energy, depending on which side it has been tuned to.

Given n crystals’ energy levels ai and types bi (1≤i≤n), bi=N means the i-th crystal is a neutral one, bi=L means a Light one, and bi=D means a Dark one. The Jedi Council asked you to choose some crystals to form a larger hybrid crystal. To make sure it is stable, the final energy level (the sum of the energy emission of all chosen crystals) of the hybrid crystal must be exactly k.

Considering the NP-Hardness of this problem, the Jedi Council puts some additional constraints to the array such that the problem is greatly simplified.

First, the Council puts a special crystal of a1=1,b1=N.

Second, the Council has arranged the other n−1 crystals in a way that

ai≤∑j=1i−1aj[bj=N]+∑j=1i−1aj[bi=L∩bj=L]+∑j=1i−1aj[bi=D∩bj=D](2≤i≤n)

[cond] evaluates to 1 if cond holds, otherwise it evaluates to 0.

For those who do not have the patience to read the problem statements, the problem asks you to find whether there exists a set S⊆{1,2,…,n} and values si for all i∈S such that

∑i∈Sai∗si=k

where si=1 if the i-th crystal is a Light one, si=−1 if the i-th crystal is a Dark one, and si∈{−1,1} if the i-th crystal is a neutral one.

Input

The first line of the input contains an integer T, denoting the number of test cases.
For each test case, the first line contains two integers n (1≤n≤10^3) and k (|k|≤10^6).
The next line contains n integer a1,a2,…,an (0≤ai ≤10^3).
The next line contains n character b1,b2,…,bn (bi ∈{L,D,N}).

Output

If there exists such a subset, output “yes”, otherwise output “no”.

Sample Input

2
5 9 1 1 2 3 4
N N N N N
6 -10 1 0 1 2 3 1
L L L L L D

Sample Output

yes
no

题目大意

有n个数，每个数都对应一个下标。这n个数的原型是a_i，当对应下标是N时，这个数可正可负；当下标是L时，这个数是正数；D时是负数。
问是否可以从这n个数中选择任意个数，使其和恰好等于k。

解题思路

我们考虑到,n的范围达到了1e3，如果要从这n个数中选择一些数直接凑的话，近乎不可能完成。
题目中又写到，为降低题目难度，增加了两个约束条件：
①.a_1一定是1，它的下标一定是N。
②.【重点】给出后n-1个数的范围：a_i ≤ （j从1——i-1的）（所有下标为N的a_j之和）+ （a_i下标为L ？ a_j下标为L的数值）+ （a_i下标为D ？ a_j下标为D的数值）。然而这个条件有什么用呢？
它间接表明第一个下标为L的数，只能取0或1；
（在不出现N时）第二个下标为L的数只能是0或1或2；
（在不出现N时）第三个下标为L的数取值范围为0-4；
（在不出现N时）第四个下标为L的数取值范围为0-8；
……
而这范围有什么用呢？
它保证了新增加下标为L的数时，整个序列可表示的数的范围的上界随之扩大，即可以用所给的数表示0-上界中的任意一个数。
（如在不出现N时前四个下标为L的a_i分别是1,2,4,8，则我可以从这些数中选出0-4个数来表示0-15中的任意一个数）
同理，把上面的L换为D，可得到整个序列能表示的数的范围的下界。
因此，在碰到下标为L的数就更新序列所能表示的数的范围的上界，碰到下标为D时就更新下界。因为下标为N的数可正可负，所以当出现N时，更新L边界的过程中就把N看作L，更新D边界过程中就把N看作D（即同时更新上下界）。
最后看k是否在能表示的范围内就可以了。

Mycode：

#include <bits/stdc++.h>using namespace std;typedef long long LL;const int MAX = 1005;const int MOD = 1e9+7;const int INF = 0x3f3f3f3f;int a[MAX];char b[MAX];int main(){    int t, n, k, uplimt, downlimt;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n, &k);        for(int i = 0; i < n; ++i)            scanf("%d", &a[i]);        for(int i = 0; i < n; ++i)            cin >> b[i];        uplimt = 1;        downlimt = -1;        for(int i = 1; i < n; ++i)        {            if(b[i] == 'N')                uplimt += a[i], downlimt -= a[i];            else if(b[i] == 'L')                uplimt += a[i];            else                downlimt -= a[i];        }        if(uplimt >= k && downlimt <= k)            puts("yes");        else            puts("no");    }    return 0;}